For $0 le theta le pi/2$, When are both $theta/pi$ and $sqrt2sintheta$ rational?
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For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?
I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.
trigonometry irrational-numbers rational-numbers
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
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add a comment |
$begingroup$
For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?
I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.
trigonometry irrational-numbers rational-numbers
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
$endgroup$
1
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
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In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
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Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
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– Jeongu Kim
Dec 5 '18 at 15:10
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Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54
add a comment |
$begingroup$
For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?
I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.
trigonometry irrational-numbers rational-numbers
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
$endgroup$
For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?
I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.
trigonometry irrational-numbers rational-numbers
trigonometry irrational-numbers rational-numbers
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
edited Dec 5 '18 at 15:08
Jeongu Kim
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
asked Dec 5 '18 at 14:52
Jeongu KimJeongu Kim
734
734
1
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10
$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54
add a comment |
1
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10
$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54
1
1
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10
$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10
$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54
add a comment |
1 Answer
1
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Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
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add a comment |
Your Answer
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 5 '18 at 23:26
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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1
$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57
$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05
$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10
$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45
$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54